Trainee - r404 boiling point

Hi all

I'm currently training to become a refrigiration engineer to repair ice cream machines, I've read "refrigiration 101" and found it very helpful.

I understand that the compressor pushes high pressured vapour through the condenser where it gains heat from the surrounding air and condenses to a liquid, it's then pushed along the lines to a expansion valve, on the other side of the expansion valve the high pressured liquid becomes low pressured and therefore has a lower boiling point, into the evaporator where it gains the heat from the surrounding area and changes to a vapour, low pressured vapour then goes back into the compressor where it all starts again,

My question is: we use R404a refrigirant. I believe the boiling point is -46 degrees. I nderstand the relationship between increasing the pressure increases the boiling point but at the evaporatornthe R404a is at 1.5 bar so not much of a pressure increase, we only need the area around the evaporator to reduce to 4 degrees, so how does the refrigirant reach the required -46 degrees to change to a vapour ? Surely the pipes etc would all reach -46 degrees too ?

Thanks for time !

Jack

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You arn't understanding the process yet.

The condenser rejects heat (not as you put it gains).

The relationship with refrigerant temperature and pressure is
if the pressure goes up the tempreature of the saturated liquid
goes up and if it falls so does the other.

You need to understand this relationship and be comfortable with it.

To correctly know what pressure and temperature the refrigerant will
be at you need a comparitor and then you will be able to compare the two.

They both change.......

At 0 barg the saturation temp of 404 is about -46degC and as the pressure
rises the saturation temp rises so at 1barg the temp would be -30 ish,
at 2barg it would be -20 ish and so on and so on.

You want a freezer to work at say -25degC so to do that the refrigerant must
boil (saturate) at a lower temp for it to work correctly. If you work to approx 10deg
differance the saturation temp will be about 10deg C colder than the air off the evap.
If the air off is -25degC then the saturation temp will be about -35degC which equals
a pressure of about 0.7barg.

If you work on machines where the cream is in contact with the barrel then the temp
differance may change but the principles remain the same.

As the evap temp changes so will the suction pressure.


Regards

Rob

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Your vocation is showing Rob!
I read this earlier and thought where to start?
Leave for someone else was my answer!
Thankfully you have responded.
Grizzly

Hi Jack,

Took me a while to figure this out back in the day. When the refrigerant is boiling at whatever temperature due to the pressure then this is the temperature of the coil (well give or take a smidgen)

So in your machine, if it is evaporating at 1.5barg (g for gauge pressure, not absolute pressure) this corresponds to a boiling point of -38*C. This means you are pumping liquid into an area of lower pressure so the liquid boils at -38*C. Yes, the coil in parts will reach that temperature as well although nearer the outlet the temperature will be a few degress higher.

Depending on the temperature of the air around the coil the saturated temperture (its boiling or condensing temperature) will rise or fall a bit. The term saturated means saturated with heat energy.

R404A boiling at -46*C is at atmospheric pressure. This means if you squirt some liquid refrigerant onto your hand, ignoring losses from conduction up your arm etc, your hand will be -46*C if you do it for long enough.

Dare you to give it a go but don't tell the F-Gas police!

Cheers,
Andy.